Problem 61 - A Graphical Approach to College Algebra

Question

Use \(f(x)\) and \(g(x)\) to find each composition. Identify its domain. (Use a calculator if necessary to find the domain.) (a) \((f \circ g)(x) \quad\) (b) \((g \circ f)(x) \quad\) (c) \((f \circ f)(x)\). $$f(x)=x^{2}, \quad g(x)=\sqrt{1-x}$$

Step-by-Step Solution

Answer

\((f \circ g)(x) = 1-x; x \leq 1.\) \((g \circ f)(x) = \sqrt{1-x^2}; -1 \leq x \leq 1.\) \((f \circ f)(x) = x^4; x \in \mathbb{R}.\)

Step 1: Understanding the Composition

The composition of functions \((f \circ g)(x)\) means we substitute \(g(x)\) into \(f(x)\). Similarly, \((g \circ f)(x)\) substitutes \(f(x)\) into \(g(x)\), and \((f \circ f)(x)\) substitutes \(f(x)\) back into itself.Let's compute each composition one by one.

Step 2: Compute \((f \circ g)(x)\)

We substitute \(g(x) = \sqrt{1-x}\) into \(f(x)\):\[(f \circ g)(x) = f(g(x)) = f(\sqrt{1-x}) = (\sqrt{1-x})^2\]This simplifies to \(1-x\).

Step 3: Domain of \((f \circ g)(x)\)

The domain of \((f \circ g)(x)\) is determined by the domain of \(g(x)\) since it is the inner function. For \(g(x) = \sqrt{1-x}\), the expression under the square root must be non-negative: \[1-x \geq 0\]Solving this gives \(x \leq 1\). Thus, the domain is \((-\infty, 1]\).

Step 4: Compute \((g \circ f)(x)\)

We substitute \(f(x) = x^2\) into \(g(x)\):\[(g \circ f)(x) = g(f(x)) = g(x^2) = \sqrt{1-x^2}\]

Step 5: Domain of \((g \circ f)(x)\)

The domain of \((g \circ f)(x)\) requires the inner function \(x^2\) not to exceed 1 under the square root:\[1 - x^2 \geq 0\]This simplifies to \(-1 \leq x \leq 1\), which is \([-1, 1]\).

Step 6: Compute \((f \circ f)(x)\)

Substitute \(f(x) = x^2\) into itself:\[(f \circ f)(x) = f(f(x)) = f(x^2) = (x^2)^2 = x^4\]

Step 7: Domain of \((f \circ f)(x)\)

Since \(f(x) = x^2\) and then \(f(f(x)) = x^4\) is defined for all real numbers, \((f \circ f)(x)\) shares the same domain as \(f(x)\).The domain is \((-\infty, \infty)\).

Key Concepts

Domain of Functions

The domain of a function is the set of all possible inputs, known as the independent variable values, for which the function is defined. It's important because it tells us the limitations of a function. When we compose functions, understanding the domain becomes crucial to ensure the composed function is valid.
When we consider a composition, like \((f \circ g)(x)\) or \((g \circ f)(x)\), the domain depends on both functions involved.

• For \((f \circ g)(x)\), it uses \(g(x)\) as the input to \(f(x)\). This means the domain of \((f \circ g)(x)\) is restricted by both the domain of \(g(x)\) and the requirement that the output of \(g(x)\) fits into the domain of \(f(x)\).
• Similarly, for \((g \circ f)(x)\), the domain is dictated first by \(f(x)\) and then the ability of \(g(x)\) to process the output from \(f(x)\).

By solving inequalities, we find that the domain of \((f \circ g)(x)\) may differ significantly from that of \(f(x)\) or \(g(x)\) alone.

Functions and Their Operations

Functions can undergo several operations: addition, subtraction, multiplication, division, and composition. These operations allow us to combine functions in various ways to form new functions.
Composition is a particular kind of operation where one function becomes the input to another. For instance, with functions \(f(x)\) and \(g(x)\), the composition \((f \circ g)(x)\) means plugging \(g(x)\) into \(f(x)\).

• Each step in the composition process affects the output based on its defined rule. The operation \((f \circ g)(x)\) results in a sequence of outputs from \(g\) that \(f\) then reshapes.
• The operation \((f \circ f)(x)\) or a self-composition involves inserting the outputs of the function into itself repeatedly.

Understanding how these operations modify the characteristics, like domain and range, of a function is vital in algebra and calculus.

Square Root Functions

Square root functions are a special type of function represented by \(\sqrt{a(x)}\), where \(a(x)\) is another function. They are unique because the expression under the square root, known as the radicand, must be non-negative for real number outputs.
The domain of a square root function is determined by solving the inequality \(a(x) \geq 0\). This restriction prevents imaginary outputs, which aren't real numbers.

• For example, in \(g(x) = \sqrt{1-x}\), we solve \(1-x \geq 0\) to find that \(x \leq 1\), which gives the domain as \(( -\infty, 1 ]\).
• Ensuring the radicand is non-negative keeps the function within the realm of real numbers, allowing the system or context using the function to perform further operations without running into undefined behavior.

Learning about square root functions helps us understand more complex compositions and interactions that involve similar constraints.